Question 6 - A-Level Maths

Edexcel F3 2024 June — Question 6 9 marks

Exam Board	Edexcel
Module	F3 (Further Pure Mathematics 3)
Year	2024
Session	June
Marks	9
Paper	Download PDF ↗
Mark scheme	Download PDF ↗
Topic	Conic sections
Type	Ellipse tangent/normal equation derivation
Difficulty	Challenging +1.3 Part (a) is a standard Further Maths ellipse tangent/normal derivation using implicit differentiation and parametric coordinates—routine for F3 students. Part (b) requires finding coordinates of Q and M, expressing triangle area in terms of θ, then optimizing using calculus, which involves several steps and trigonometric manipulation but follows established techniques without requiring novel geometric insight.
Spec	1.07m Tangents and normals: gradient and equations 1.07r Chain rule: dy/dx = dy/du * du/dx and connected rates

The ellipse $E$ has equation

$$\frac { x ^ { 2 } } { 25 } + \frac { y ^ { 2 } } { 9 } = 1$$ The line $l$ is the normal to $E$ at the point $P ( 5 \cos \theta , 3 \sin \theta )$ where $0 < \theta < \frac { \pi } { 2 }$

Using calculus, show that an equation for $l$ is $$5 x \sin \theta - 3 y \cos \theta = 16 \sin \theta \cos \theta$$ Given that
- $\quad l$ intersects the $y$-axis at the point $Q$
- the midpoint of the line segment $P Q$ is $M$
- determine the exact maximum area of triangle $O M P$ as $\theta$ varies, where $O$ is the origin.
You must justify your answer.

Show mark scheme Show mark scheme source

Question 6(a):

Answer	Marks	Guidance
Answer/Working	Mark	Guidance
$\frac{dx}{d\theta} = -5\sin\theta$, $\frac{dy}{d\theta} = 3\cos\theta$ giving $\frac{dy}{dx} = -\frac{3\cos\theta}{5\sin\theta}$	B1	Any correct expression for $\frac{dy}{dx}$ in terms of $\theta$, or $x$ and $y$, or $x$
$m_T = -\frac{3\cos\theta}{5\sin\theta} \Rightarrow m_N = \frac{5\sin\theta}{3\cos\theta}$	M1	Correct perpendicular gradient rule for their $\frac{dy}{dx}$ in terms of $\theta$. May see $m_T = -\frac{3}{5}\cot\theta \Rightarrow m_N = \frac{5}{3}\tan\theta$
$y - 3\sin\theta = \frac{5\sin\theta}{3\cos\theta}(x - 5\cos\theta)$ OR $c = -\frac{16}{3}\sin\theta$	M1	Correct straight line method with a changed gradient in terms of $\theta$
$3y\cos\theta - 9\sin\theta\cos\theta = 5x\sin\theta - 25\sin\theta\cos\theta$ $\Rightarrow 5x\sin\theta - 3y\cos\theta = 16\sin\theta\cos\theta$*	A1*	Reaches given answer with intermediate working. Allow $x$ and $y$ terms in different order provided together, third term on other side, products in different order, coefficients "5", "–3" and "16" at front

Question 6(b):

Answer	Marks	Guidance
Answer/Working	Mark	Guidance
At $Q$, $x=0 \Rightarrow y = -\frac{16}{3}\sin\theta$	B1	Correct $y$ coordinate of $Q$. Accept unsimplified
$M$ is $\left(\frac{5\cos\theta}{2}, \frac{3\sin\theta - \frac{16}{3}\sin\theta}{2}\right)$, accept $x = \frac{5}{2}\cos\theta$, $y = -\frac{7}{6}\sin\theta$	M1	Correct method for midpoint for both coordinates with their $y_Q$. Could be implied
Area $\Delta OPM = \frac{1}{2} \times \frac{16}{5}\cos\theta\left(3\sin\theta + \frac{7}{6}\sin\theta\right)$	M1	Correct unsimplified expression for area of $\Delta OPM$. Do not allow recovery from negative area. Can only follow incorrect work if $\Delta OPM = \frac{1}{2}\Delta OPQ$ is used
$\left\{=\frac{20}{3}\sin\theta\cos\theta = \frac{10}{3}\sin 2\theta\right\} \Rightarrow \text{area} = \frac{10}{3}$	A1	Correct area following a correct expression
$\frac{10}{3}$ and justification: max of $\sin 2\theta$ is 1, or $\theta = \frac{\pi}{4}$ or 45°, or differentiates $\frac{10}{3}\sin 2\theta \Rightarrow \frac{20}{3}\cos 2\theta = 0$	A1	Do not accept wrong statements e.g. $\sin 2\theta \leq 1$, $-1 < \sin 2\theta < 1$ but condone ambiguous "$\sin 2\theta$ is between 1 and –1". From any other expression must differentiate unless rewritten as $\frac{10}{3}\sin 2\theta$

# Question 6(a):

| Answer/Working | Mark | Guidance |
|---|---|---|
| $\frac{dx}{d\theta} = -5\sin\theta$, $\frac{dy}{d\theta} = 3\cos\theta$ giving $\frac{dy}{dx} = -\frac{3\cos\theta}{5\sin\theta}$ | B1 | Any correct expression for $\frac{dy}{dx}$ in terms of $\theta$, or $x$ and $y$, or $x$ |
| $m_T = -\frac{3\cos\theta}{5\sin\theta} \Rightarrow m_N = \frac{5\sin\theta}{3\cos\theta}$ | M1 | Correct perpendicular gradient rule for their $\frac{dy}{dx}$ in terms of $\theta$. May see $m_T = -\frac{3}{5}\cot\theta \Rightarrow m_N = \frac{5}{3}\tan\theta$ |
| $y - 3\sin\theta = \frac{5\sin\theta}{3\cos\theta}(x - 5\cos\theta)$ OR $c = -\frac{16}{3}\sin\theta$ | M1 | Correct straight line method with a changed gradient in terms of $\theta$ |
| $3y\cos\theta - 9\sin\theta\cos\theta = 5x\sin\theta - 25\sin\theta\cos\theta$ $\Rightarrow 5x\sin\theta - 3y\cos\theta = 16\sin\theta\cos\theta$* | A1* | Reaches given answer with intermediate working. Allow $x$ and $y$ terms in different order provided together, third term on other side, products in different order, coefficients "5", "–3" and "16" at front |

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# Question 6(b):

| Answer/Working | Mark | Guidance |
|---|---|---|
| At $Q$, $x=0 \Rightarrow y = -\frac{16}{3}\sin\theta$ | B1 | Correct $y$ coordinate of $Q$. Accept unsimplified |
| $M$ is $\left(\frac{5\cos\theta}{2}, \frac{3\sin\theta - \frac{16}{3}\sin\theta}{2}\right)$, accept $x = \frac{5}{2}\cos\theta$, $y = -\frac{7}{6}\sin\theta$ | M1 | Correct method for midpoint for both coordinates with their $y_Q$. Could be implied |
| Area $\Delta OPM = \frac{1}{2} \times \frac{16}{5}\cos\theta\left(3\sin\theta + \frac{7}{6}\sin\theta\right)$ | M1 | Correct unsimplified expression for area of $\Delta OPM$. Do not allow recovery from negative area. Can only follow incorrect work if $\Delta OPM = \frac{1}{2}\Delta OPQ$ is used |
| $\left\{=\frac{20}{3}\sin\theta\cos\theta = \frac{10}{3}\sin 2\theta\right\} \Rightarrow \text{area} = \frac{10}{3}$ | A1 | Correct area following a correct expression |
| $\frac{10}{3}$ and justification: max of $\sin 2\theta$ is 1, or $\theta = \frac{\pi}{4}$ or 45°, or differentiates $\frac{10}{3}\sin 2\theta \Rightarrow \frac{20}{3}\cos 2\theta = 0$ | A1 | Do not accept wrong statements e.g. $\sin 2\theta \leq 1$, $-1 < \sin 2\theta < 1$ but condone ambiguous "$\sin 2\theta$ is between 1 and –1". From any other expression must differentiate unless rewritten as $\frac{10}{3}\sin 2\theta$ |

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Show LaTeX source

\begin{enumerate}
  \item The ellipse $E$ has equation
\end{enumerate}

$$\frac { x ^ { 2 } } { 25 } + \frac { y ^ { 2 } } { 9 } = 1$$

The line $l$ is the normal to $E$ at the point $P ( 5 \cos \theta , 3 \sin \theta )$ where $0 < \theta < \frac { \pi } { 2 }$\\
(a) Using calculus, show that an equation for $l$ is

$$5 x \sin \theta - 3 y \cos \theta = 16 \sin \theta \cos \theta$$

Given that

\begin{itemize}
  \item $\quad l$ intersects the $y$-axis at the point $Q$
  \item the midpoint of the line segment $P Q$ is $M$\\
(b) determine the exact maximum area of triangle $O M P$ as $\theta$ varies, where $O$ is the origin.
\end{itemize}

You must justify your answer.

\hfill \mbox{\textit{Edexcel F3 2024 Q6 [9]}}

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